1.
a in T, there exists n such that a^n = e
b in T, there exists m such that b^m = e
(ab)^mn = (a^n)^m (b^m)^n = e^m e^n = e
axiom 1 holds.
2.
let e be identiy in G
e^n = e ==> e in T(G)
T(G) subgroup of G, it's also true that ea = ae =a for all a in G
conclusion: idenity of G, e...
Homework Statement
Homework Equations
subgroup axioms:
1. a, b in T(G), then ab in T(G)
2. existence of identity element.
3. a in T(G), then a^-1 in T(G)
The Attempt at a Solution
1.
let a be in T(G), then a^n = e.
let b be in T(G), then b^n = e
(ab)^n = (a^n)(b^n) = (e)(e) = e
axiom 1...
My fault all along. I was trying to calculate the magnitude of the surface instead of the magnitude of the gradient of the surface.
Yes i know how to calculate the magnitude of a vector thank you very much.
answered.
Homework Statement
Find vector normal to z = x^2 + y^2 - 3 at point r = (2, -1, 2)
Homework Equations
The Attempt at a Solution
here is the markscheme. I understand how to find the gradient, but i dont understand how they calculated the magnitude.
thanks
Homework Statement
Let V be a vector space over a field F and let L and M be two linear transformations from V to V.
Show that the subset W := {x in V : L(x) = M(x)} is a subspace of V .
The Attempt at a Solution
I presume it's a simple question, but it's one of those where you just don't...
I think i understand.
L(1) means that we are considering the function f(x) = 1 (a straight line through y=1):
L(f) = f' + f(-2)t
L(1) = d(1)/dt + f(-2)t
f(-2) is equal to 1 in this case, so:
d(1)/dt + f(-2)t = 0 +t = t
However, if we consider L(t):
L(t) = t' + f(-2)t
=d(t)/dt + f(-2)t...
Homework Statement
Here is the question, i know how to do part (i) but I do not understand part (ii):
The Attempt at a Solution
[/B]
here's the solution from the marking scheme:
i understand how they formed the matrix from their working out (i can se the pattern), but I do not...
Homework Statement
[/B]
find directional derivative at point (0,0) in direction u = (1, -1) for
f(x,y) = x(1+y)^-1
The Attempt at a Solution
grad f(x,y) = ( (1+y)^-1, -x(1+y)^-2 )
grad f(0,0) = (1, 0)
grad f(x,y) . u = (1,0).(1,-1) = 1.
seems easy but markscheme says im wromg. It says...